This is the answer to the Add Two Numbers - LeetCode problem. Please don’t use it; the solution is very very inefficient. Here are some reasons this code sucks:
- Do three separate pass throughs on l1, l2 and creating a linked list from their integer sum
digit.val * (radix ** index_a) is a very expensive operation.
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52 | # Definition for singly-linked list.
# class ListNode:
# def __init__(self, val=0, next=None):
# self.val = val
# self.next = next
class Solution:
def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode:
radix = 10
running_sum = 0
# iteratively sum l1
digit = l1
index_a = 0
while digit:
running_sum += digit.val * (radix ** index_a)
digit = digit.next
index_a += 1
l1_sum = running_sum
# iteratively sum l2's digits to l1 for running_sum
digit = l2
index_b = 0
while digit:
running_sum += digit.val * (radix ** index_b)
digit = digit.next
index_b += 1
total_sum = running_sum
ret_list = None
if not running_sum >= (radix ** index_a) or not running_sum >= (radix ** index_b):
index = max(index_a-1, index_b-1)
else:
index = max(index_a, index_b)
# create the return list
while index >= 0:
digit = total_sum // (radix ** index)
print(total_sum, '//', (radix ** index), '=', digit)
total_sum = total_sum % (radix ** index)
index -= 1
ret_list_old = ret_list
ret_list = ListNode(digit)
ret_list.next = ret_list_old
return ret_list
|
Like this post? Share on:
Twitter
❄ Facebook
❄ Email